Answer:
To find critical points, take the first derivative and set it equal to zero:
f(x) = -2x^2 + 4x + 5
f'(x) = -4x + 4
-4x+4 = 0
-4x = -4
x = 1
Critical point at x = 1
Alternatively, if you mean zeros, or where the x intersects, you can use the quadratic equation.
Answer:
112
Step-by-step explanation:
3(7-3) = 3 x 4 = 12
12^2 = 144
144 -4(6+2)
144-4(8)
144-32
112
Answer: 350
Step-by-step explanation:
Answer:
40.56
Step-by-step explanation:
2pir
8÷2=r=4
2pi4=25.13
25.13÷2=12.565
10+10+8+12.56=40.565